why inductor voltage leads source voltage

[beenuseren]

If we the voltage waveforms of an inductor and the source voltage, its clear that the inductor voltage leads the source voltage.

When does the voltage across the inductor starts reducing????
It reduces when it starts discharging. And when can it discharge???
Only when the supply voltage is lower than the inductor voltage.

But if we see the wave forms, the inductor voltage is reducing even when the supply voltage is comparitively above and increasing..

How is it possible????
Please refer to the waveform diagram I have attached.

Thankyou.

 

[Boncuk]

Hi,

an inductor is only a temporarily charging device. As soon as the supply voltage decreases - no matter what the inductor's voltage is - the inductor will "discharge" by changing the magnetic field, thereby producing voltage itself.

Observe the current flow as well. It has a phase shift of 90deg compared with the voltage.

Boncuk

 

[duffy]

An inductor is like a rock. On a rollerskate. You give it a shove, and at first it doesn't move, then it gets going, then it wants to keep going even when you are pulling it backwards.

How is it possible????

Nobody even understands the rock.

 

[MrAl]

Hi,

You asked this question in another thread and i gave you this answer:

Thanks MRAI,

What you said is right. The capacitor voltage lags the supply voltage.

Another doubt, the same wave forms hold good for a RL circuit.
My previous questions apply for this too.
How come the inductor voltage starts reducing, when the supply voltage is increasing and above the inductor voltage???

Please check the waveform I have attached...

Hi again,


That's a good question really, as the inductor voltage would lead
which looks a little strange at first, but only if you think of the
inductor as completely 'passive'.

The inductor is considered passive in some types calculations,
but for this kind of question the inductor has to be thought of as
an energy storage element, which is capable of delivering energy
as well as receiving it. It's when the inductor is delivering energy
that we see that it is definitely not passive, but active in that
it can supply current and voltage to a circuit just like a battery
would, at least once it has some energy stored in it.
The cap does this too, only it expels its energy in the form of
a current, while the inductor does so in the form of a voltage
most basically, as when it is trying to maintain a constant
current it forces the voltage to some value to satisfy this
condition. We could also look at the cap current and ask
a few questions too.

To best understand the inductor over time we have to look at
this little equation:

v=L*di/dt

where

v is the voltage across the inductor
L is the inductance value
di is the change in current over time increment dt
dt is the change in time.

Note one thing here: that the voltage (which the waveform shows)
depends not on the supply excitation voltage, but the value of the
inductance L and the rate of change of the current di/dt.
Thus, it is the rate of change of current that determines the voltage,
not the supply voltage by itself.

What this means is that if you draw the current wave and then look
at the inductor voltage wave and think about the derivative of the
current wave you will see the inductor voltage follow this relationship.
Where the peaks of the current are flat the derivative is zero, so
the voltage will pass through zero at those points, and where the
current wave is changing the fastest, the current will reach its peaks.

Another way is to simply plot the derivative of the current, multiply
by the inductance L, and there you have the exact voltage wave.

Oh yeah, BTW, it's unfortunate that some fonts render an upper case
"i" as l as well as lower case "L", but with a better font this is my
screen name:

MrAl

or in all upper case:

MRAL

This is a problem that comes up now and then.

 

[Hero999]

First of all forget the sine wave.

Connect the inductor to a DC supply in series with a resistor.

Watch the voltage and current across the inductor. When you first connect the power the voltage will be equal to the supply and the current will be zero. As the inductor magnetises the current will rise to what the resistor permits and the voltage across the inductor will fall to zero.

The same is true for an AC wavefrom, if the inductor is perfect the voltage will lead the current by 90° and the power dissipated in the inductor will be zero.